Answer:
Explanation:
We shall find out the molecular formula of the substance .
Ration of number of atoms of C , O and H
=
= 3.12 : 3.12 : 12.58
= 1 : 1 : 4
volume of gas at NTP
= 250 x 273 / 350 mL .
= 195 mL .
Molecular weight of the substance = .2804 x 22400 / 195 g
= 32. approx
Let the molecular formula be
(COH₄)n
n x 32 = 32
n = 1
Molecular formula = COH₄
The compound appears to be CH₃OH
a )
CO + 2H₂ = CH₃OH
28g 4g 32g
59 8
For 8 kg hydrogen , CO required = 56 kg
CO is in excess . hydrogen is the limiting reagent .
mass of product formed
= 32 x 8 / 4
= 64 kg
b )
percentage yield = product actually formed / product to be formed theoretically x 100
= 59.6 x 100 / 64
= 93.12 %
c )
2CH₃OH + 3O₂ = 2CO₂ + 4H₂O .
64 g 2 x 22.4 L
Gram of gas in 1 gallon of fuel
= .7914 x 3785
= 2995.5 g
CO₂ produced at NTP by 2995.5 g CH₃OH
= 2 x 22.4 x 2995.5 / 64 L
= 2096.85 L
At 27° C and 766 mm Hg , this volume is equal to
2096.85 x 300 x 760 / 273 x 766
= 2286.18 L .
d )
C₈H₁₈ = 8CO₂
114g 8 x 22.4 L
gram of fuel per unit gallon
= .6986 x 3785
= 2644.2g
gram of CO₂ produced by 1 gallon of fuel at NTP
= 8 x 22.4 x 2644.2 / 114
= 4156.5 L
So it produces more CO₂ .